Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Locus Solutions Exercise 1(a) will help students to clear their doubts quickly.

## Intermediate 1st Year Maths 1B Locus Solutions Exercise 1(a)

I.

Question 1.

Find the equation of the locus of a point that is at a distance 5 from A (4, – 3).

Solution:

A (4, -3) is the given point P(x, y) is any point on the locus.

Given condition is CP = 5

CP² = 25

(x – 4)² + (y + 3)² = 25

x² – 8x + 16 + y² + 6y + 9 – 25 = 0

Equation of the locus of P is x²+ y² – 8x + 6y = 0.

Question 2.

Find the equation of locus of a point which is equidistant from the points A(-3, 2) and B(0, 4).

Solution:

A (-3, 2), B (0,4) are the given points

P (x, y) is any point on the locus

Given condition is PA = PB

PA² = PB²

(x+3)² + (y – 2)² = (x – 0)² + (y – 4)²

x² + 6x + 9 + y² – 4y + 4 = x² + y² – 8y +16

6x + 4y = 3 is the equation of the locus.

Question 3.

Find the equation of locus of a point P such that the distance of P from the origin is twice the distance of P from A (1, 2).

Solution:

O(0,0), A (1,2) are the given points

P (x, y) is any point on the locus

Given condition is OP = 2AP

OP² = 4 AP²

x² + y² = 4 [(x – 1)² + (y – 2)²]

= 4 (x² – 2x + 1 + y² – 4y + 4)

x² + y² = 4x² + 4y² – 8x – 16y + 20

Equation to the locus of P is

3x² + 3y² – 8x – 16y + 20 = 0.

Question 4.

Find the equation of locus of a point which is equidistant from the coordi¬nate axes.

Solution:

P(x, y) is any point on the locus.

PM and PN are perpendiculars from P on X and Y – axes.

Given PM = PN ⇒ PM² = PN²

y² = x²

Locus of P is x² – y² = 0

Question 5.

Find the equation of locus of a point equidistant from A (2, 0) and the Y- axis.

Solution:

A (2,0) is the given point.

P (x, y) is any point on the locus.

Draw PN perpendicular to Y – axis.

Given condition is PA = PN

PA² = PN²

(x – 2)² + (y – 0)² = x²

x² – 4x + 4 + y² = x²

Locus of P is y² – 4x + 4 = 0

Question 6.

Find the equation of locus of a point P, the square of whose distance from the origin is 4 times its y coordinate.

Solution:

P(x, y) is any point on the locus

Given condition is OP² = 4y ⇒ x² + y² = 4y

Equation of the locus of P is x² + y² – 4y = 0

Question 7.

Find the equation of locus of a point ‘P’ such that PA² + PB² = 2c², where A = (a, 0), B = (-a, 0) and 0 < |a| < |c|.

Solution:

Let P (x, y) be a point in locus.

A = (a, 0); B = (-a, 0)

Given condition is PA² + PB² = 2c²

(x – a)² + (y – 0)² + (x + a)² + (y – 0)² = 2c²

x² – 2ax + a² + y² + x² + 2ax + a² + y² = 2c²

2x² + 2y² = 2c² – 2a²

∴ x² + y² = c² – a² is the locus.

II.

Question 1.

Find the equation of locus of P, if the line segment joining (2, 3) and (-1, 5) subtends a right angle at P.

Solution:

A(2,3), B (-1, 5) are the given points.

P(x, y) is any point on the locus.

Given condition is, ∠APB = 90°

AP² + PB² = AB²

(x – 2)² + (y – 3)² + (x + 1)² + (y – 5)² = (2 + 1)² + (3 – 5)²

x² – 4x + 4 + y² – 6y + 9 + x² + 2x + 1+ y² – 10y + 25 = 9 + 4

2x² + 2y² – 2x – 16y + 26 = 0

Locus of P is x² + y² – x – 8y + 13 = 0

(x, y) ≠ (2,3) and (x, y) ≠ (-1, 5)

Question 2.

The ends of the hypotenuse of a right angled triangle are (0, 6) and (6, 0). Find the equation of locus of its third vertex.

Solution:

A(0,6), B (6,0) are the ends of the hypotenuse,

P (x, y) is the third vertex.

∴ The given condition is ∠APB = 90°

AP² + PB² = AB²

(x – 0)² + (y – 6)² + (x – 6)² + (y – 0)² = (6 – 0)² + (0 – 6)²

x² + y² – 12y + 36 + x² – 12x + 36 + y² = 36 + 36

2x² + 2y² -12x -12y = 0

Locus of P is x² + y² – 6x – 6y = 0

(x, y) ≠ (0,6) and (x, y) ≠ (6,0).

Question 3.

Find the equation of the locus of a point, the difference of whose distances from (-5, 0) and (5, 0) is 8.

Solution:

A(5,0), B(-5,0) are the given points.

P(x, y) is any point on the locus.

Given condition is |PA – PB| = 8

PA – PB = 8 ………..(1)

PA² – PB² = [(x – 5)² + (y – 0)²] – [(x + 5)² + (y – 0)²]

= x² – 10x + 25 + y² – x² – 10x – 25 – y²

= – 20x

(PA + PB) (PA – PB) = -20x

(PA + PB) 8 = -20x

PA + PB = –\(\frac{5}{2}\)x …………..(2)

Adding (1) and (2),

2PA = –\(\frac{5x}{2}\) + 8 = \(\frac{-5 x+16}{2}\)

4PA = – 5x + 16

16PA² = (- 5x + 16)²

16 [(x – 5)² + y²] = (- 5x + 16)²

16 [x² – 10x + 25 + y²] = [- 5x + 16]²

16x² + 16y²2 – 160x + 400 = 25x² + 256 – 160x

9x² – 16y² = 144

Dividing with 144, locus of P is

\(\frac{9 x^{2}}{144}-\frac{16 y^{2}}{144}=1\) i.e., \(\frac{x^{2}}{16}-\frac{y^{2}}{9}=1\)

Question 4.

Find the equation of locus Of P, if A=(4,0), B = (- 4,0) and |PA- PB| =4.

Solution:

A = (4, 0), B = (-4, 0) are the given points.

P (x, y) is any point on the locus.

The given condition is |PA – PB| = 4 …………… (1)

PA² – PB²= [(x- 4)² + (y – 0)²] – [(x + 4)² + y²]

= x² – 8x+ 16 + y² – x² – 8x – 16 – y²

= -16x

(PA + PB) (PA – PB) = -16x

(PA + PB) 4 = – 16x

PA + PB = – 4x ………….. (2)

Adding (1) and (2),

2PA = 4 – 4x

PA = 2 – 2x

PA² = (2 – 2x)²

(x – 4)² + (y – 0)² = (2 – 2x)²

x² – 8x + 16 + y² = 4 + 4x² – 8x

3x² – y² = 12

Dividing with 12, locus of P is \(\frac{3 x^{2}}{12}-\frac{y^{2}}{12}=1\)

i.e., \(\frac{x^{2}}{4}-\frac{y^{2}}{13}=1\)

Question 5.

Find the equation of the locus of a point, the sum of whose distances from (0, 2) and (0, -2) is 6.

Solution:

A (0,2), B (0, -2) are the given points.

P(x, y) is any point on the locus.

Given condition is PA + PB = 6 ……….. (1)

PA² – PB² = [(x – 0)² + (y – 2)²] – [(x – 0)² + (y + 2)²]

= x² + y² – 4y + 4 – x² – y² – 4y – 4 = – 8y

(PA + PB) (PA – PB) = -8y

6(PA – PB) = – 8y

PA – PB = –\(\frac{8y}{6}\)

PA-PB = –\(\frac{4y}{3}\) ……. (2)

Adding (1) and (2), 2PA = 6 –\(\frac{4y}{3}\)

9x² + 9y² + 36 = 81 + 4y²

9x² + 5y² = 45

Dividing with 45,

Question 6.

Find the equation (rf the locus of P, if A= (2,3), B = (2, -3) and PA + PB = 8.

Solution:

A (2, 3), B (2, -3) are the given points.

P(x, y) is any point on the locus.

Given condition is PA + PB = 8 ……….. (1)

PA² – PB² = [(x – 2)² + (y – 3)²] – [(x – 2)² + (y + 3)²]

= (x – 2)² + (y – 3)² – (x – 2)² – (y + 3)² = (y – 3)² – (y + 3)² = -12y

(PA + PB) (PA – PB) = -12y

8(PA – PB) = -12y

PA – PB = \(\frac{-12 y}{8}\)

PA – PB = \(\frac{-3 y}{2}\) ……….. (2)

Adding (1) and (2),

2PA = 8 – \(\frac{3 y}{2}\) = \(\frac{16 -3y}{2}\)

4PA = 16 – 3y

16PA² = (16 – 3y)²

16 [(x – 2)² + (y – 3)²] = (16 – 3y)²

16(x² – 4x + 4 + y² – 6y + 9) = (16 – 3y)²

16x² + 16y² – 64x – 96y + 208 = 256 + 9y² – 96y

16x² + 7y² – 64x – 48 = 0

Locus of P is 16x² + 7y² – 64x – 48 = 0.

Question 7.

A(5, 3) and B (3, -2) are two fixed points. Find the equation of the locus of P, so that the area of triangle PAB is 9.

Solution:

A(5, 3), B(3, -2) are the given points.

P(x, y) is any point on the locus.

Given condition is ∆PAB = 9

\(\frac{1}{2}\)|5(-2, -y) + 3(y – 3) + x(3 + 2)|=9

|-10 – 5y + 3y – 9 + 5x| = 18

5x – 2y – 19 = ± 18

5x – 2y – 19 = 18 or 5x – 2y – 19 = – 18

5x – 2y – 37 = 0 or 5x – 2y – 1 = 0

Locus of P is (5x- 2y- 37) (5x- 2y- 1) = 0

Question 8.

Find the equation of the locus of a point, which forms a triangle of ar5a 2 with the points A(l, 1) and B (-2, 3).

Solution:

A (1, 1), B(-2, 3) are the given points.

P(x, y) is any point on the locus.

Given condition is ∆PAB = 2

\(\frac{1}{2}\)|1(3 – y) – 2 (y – 1) + x (1 – 3)| = 2

|3 – y – 2y + 2 – 2x| = 4

-2x – 3y + 5 = ± 4

-2x – 3y + 5 = 4 or – 2x – 3y + 5 = – 4

2x + 3y – 1 = 0 or 2x + 3y – 9 = 0

Locus of P is (2x + 3y – 1) (2x + 3y – 9) = 0.

Question 9.

If the distance from ‘P’ to the points (2, 3) and (2, -3) are in the ratio 2 : 3, then find the equation of locus of P.

Solution:

Let P (x, y) be a point on locus.

Given points A = (2, 3), B = (2, -3)

Given condition is

PA : PB = 2 : 3

⇒ 3PA = 2PB

⇒ 9PA² = 4PB²

⇒ 9[(x – 2)² + (y – 3)²] = 4[(x – 2)² + (y + 3)²]

⇒ 9[x² – 4x + 4 + y² – 6y + 9] = 4 [x² – 4x + 4 + y² + 6y + 9]

∴ 5x² + 5y² – 20 x – 78 y + 65 = 0 is the equation of locus.

Question 10.

A (1, 2), B (2, -3) and C (-2, 3) are three points. A point ‘P’ moves such that PA² + PB² = 2PC². Show that the equation to the locus of ‘P’ is 7x – 7y + 4 = 0.

Solution:

Let P (x, y) be a point on locus.

Given points A = (1, 2), B = (2) -3) and C = (-2, 3)

Given condition is PA² + PB² = 2 PC²

⇒ (x – 1)² + (y – 2)² + (x – 2)² + (y + 3)² = 2 [(x + 2)² + (y-3)²]

⇒ 2x² + 2y² – 6x + 2y + 18 = 2x² + 2y² + 8x – 12y + 26

⇒ 14x – 14y + 8 = 0

∴ 7x – 7y + 4 = 0 is the equation of locus.

SOLVED PROBLEMS

Question 1.

Find the equation of the locus of a point which is at a distance 5 from (-2, 3) in the xoy plane.

Solution:

Let the given point be A = (-2, 3) and P(x, y) be a point on the plane.

The geometric condition to be satisfied by P to be on the locus is that

AP = 5 …………… (1)

Expressing this condition algebraically, we get

\(\sqrt{(x+2)^{2}+(y-3)^{2}}=5\)

i.e., x² + 4x + 4 + y² – 6y + 9 = 25

i.e., x² + y² + 4x – 6y – 12 = 0 ……………. (2)

Let Q(x_{1}, y_{1}) satisfy (2).

Hence AQ = 5.

This means that Q(x_{1}, y_{1}) satisfies the geometric condition (1).

∴ The required equation of locus is

x² + y² + 4x – 6y – 12 = 0.

Question 2.

Find the equation of locus of a point P, if the distance of P from A(3,0) is twice the distance of P from B(-3, 0).

Solution:

Let P(x, y) be a point on the locus. Then the geometric condition to be satisfied by P is

PA = 2PB …………. (1)

i.e., PA² = 4PB²

i.e., (x – 3)² + y² = 4 [(x + 3)² + y²]

i.e., x² – 6x + 9 + y² = 4 [x² + 6x + 9 + y²]

i.e., 3x² + 3y² + 30x + 27 = 0

i.e., x2 + y2 + lOx + 9 = 0 ………….. (2)

Let Q(x_{1}, y_{1}) satisfy (2).

∴ QA = 2QB.

This means that Q(x_{1}, y_{1}) satisfies (1).

Hence, the required equation of locus is

x² + y² + 10x + 9 = 0.

Question 3.

Find the locus of the third vertex of a right angled triangle, the ends of whose hypotenuse are (4, 0) and (0, 4).

Solution:

Let A = (4, 0) and B = (0, 4).

Let P(x, y) be a point such that PA and PB are

perpendicular. Then PA² + PB² = AB².

i.e., (x – 4)² + y² + x² + (y – 4)² = 16 + 16

i.e., 2x² + 2y² – 8x – 8y = 0

or x² + y² – 4x – 4y = 0

Let Q(x_{1}, y_{1}) satisfy (2) and Q be different from A and B.

Hence QA² + QB² = AB², Q ≠ A and Q ≠ B.

This means that Q(x_{1}, y_{1}) satisfies (1).

∴ The required equation of locus is (2), which is the circle with \(\overline{\mathrm{AB}}\) as diameter, deleting the points A and B.

Though A and B satisfy equation (2), they do not satisfy the required geometric condition.

Question 4.

Find the equation of the locus of P, if the ratio of the distances from P to A(5, -4) and B(7, 6) is 2 : 3.

Solution:

Let P(x, y) be any point on the locus.

The geometric condition to be satisfied by P

is\(\frac{AP}{PB}\) = \(\frac{2}{3}\).

i.e., 3AP = 2PB ………. (1)

i.e., 9AP² = 4PB²

i.e., 9[(x – 5)² + (y + 4)²] = 4[(x – 7)² + (y – 6)²]

i.e., 9[x² + 25 – 10x + y² + 16 + 8y] = 4[x² + 49 – 14x +y² + 36 – 12y]

i.e, 5x² + 5y² – 34x + 120y + 29 = 0 ……… (2)

Let Q(x_{1} y_{1}) satisfy (2). Then

(by using (3))

– 4 [(x_{1} – 7)² + (y_{1} – 6)²] = 4PB²

Thus 3AQ = 2PB. This means that Q(x_{1}, y_{1}) satisfies (1).

Hence, the required equation of locus is

5(x² + y²) – 34x + 120y + 29 = 0.

Question 5.

A(2, 3) and B(-3, 4) are two given points. Find the equation of locus of P so that the area of the triangle PAB is 8.5.

Solution:

Let P(x, y) be a point on the locus.

The geometric condition to be satisfied by P is that.

area of ∆PAB = 8.5 ……….. (1)

i.e., \(\frac{1}{2}\)|x(3 – 4) + 2(4 – y) – 3(y – 3)| = 8.5

i.e., |-x + 8-2y-3y+9| = 17

i.e., |-x- 5y + 17| = 17

i.e., -x- 5y + 17 = 17 or-x-5y + 17 = -17

i.e., x + 5y = 0 or x + 5y = 34

∴ (x + 5y) (x + 5y – 34) = 0

i.e., x² + 10xy + 25y² – 34x – 170y = 0 …………. (2)

Let Q(x_{1}, y_{1}) satisfy (2). Then

x_{1} + 5y_{1} = 0 or x_{1} + 5y_{1} = 34 ………… (3)

Now, area of ∆QAB

= \(\frac{1}{2}\)|x_{1}(3-4) + 2(4-y_{1}) – 3(y_{1} – 3)|

= \(\frac{1}{2}\) |-x_{1} + 8 – 2y_{1} – 3y_{1} + 9|

= \(\frac{1}{2}\) |-x_{1} – 5y_{1} + 17|

= \(\frac{17}{2}\) =8.5 (by using (3))

This means that Q(x_{1}, y_{1}) satisfies (1).

Hence, the required equation of locus is

(x + 5y) (x + 5y – 34) = 0 or

x² + 10xy + 25 – y² – 34x- 170y = 0.